Q. If M is a metric space such that every infinite subset has a limit point, then show that the space is separable.
I want to show that we can cover M with finitely many open balls with a fixed radius $\epsilon$ and that should work for any radius $\epsilon > 0$. But how do I construct a countable dense subset?
P.S. I referred to other solutions to the same question but couldn't understand the mesh concept.
$(M, d) $ be a metric space and every infinite subsets of $M$ has a limit point, then $(M, d) $is compact (limit point compact, for a metric space limit point compact,sequentially compact,and open cover compact are all equivalent) . Choose, an open cover $\{B(x, 1/n) : x\in M \}$ . Then by Compactness, it admits a finite subcover $\{B(x_i, 1/n) :x_i\in S\}$ , $S\subset M$ finite. Then the set $S=\{x_i ;i\in I\}$ Index set $I$is finite, is countable dense subset of $(M, d)$.To prove dense, choose any point $x\in M$, $x\in B(x_i, 1/n) $ for some$ x_i\in S$, then $d(x, x_i) <1/n$.